[R] How to visualise what code is processed within a for loop
Rui Barradas
ru|pb@rr@d@@ @end|ng |rom @@po@pt
Mon Apr 30 17:56:49 CEST 2018
Hello,
Another thing to note is that regexpr is likely to take (much) more time
than ifelse or as.integer.
And the code will therefore not be very optimizable.
Rui Barradas
On 4/30/2018 4:25 PM, MacQueen, Don wrote:
> Luca,
>
> If speed is important, you might improve performance by making d0 into a
> true matrix, rather than a data frame (assuming d0 is indeed a data
> frame at this point). Although data frames may look like matrices, they
> aren’t, and they have some overhead that matrices don’t. I don’t think
> you would be able to use the [[nm]] syntax with a matrix, but [ , nm]
> should work, provided the matrix has column names. Or you could perhaps
> index by column number.
>
> I had a project some years ago in which I reduced calculation time a lot
> by extracting the numeric columns of a data frame and working with them,
> then recombining them with the character columns. R’s performance
> working with data frames has improved since then, so I really don’t know
> if it would make a difference for your task.
>
> -Don
>
> --
>
> Don MacQueen
>
> Lawrence Livermore National Laboratory
>
> 7000 East Ave., L-627
>
> Livermore, CA 94550
>
> 925-423-1062
>
> Lab cell 925-724-7509
>
> *From: *Luca Meyer <lucam1968 using gmail.com>
> *Date: *Monday, April 30, 2018 at 8:08 AM
> *To: *Rui Barradas <ruipbarradas using sapo.pt>
> *Cc: *"MacQueen, Don" <macqueen1 using llnl.gov>, array R-help
> <r-help using r-project.org>
> *Subject: *Re: [R] How to visualise what code is processed within a for loop
>
> Hi Rui
>
> Thank you for your suggestion,
>
> I have tested the code suggested by you against that supplied by Don in
> terms of timing and results are very much aligned: to populate a
> 5954x899 0/1 matrix on my machine your procedure took 79 secs, while the
> one with ifelse employed 80 secs, hence unfortunately not really any
> significant time saved there.
>
> Nevertheless thank you for your contribution.
>
> Kind regards,
>
> Luca
>
> 2018-04-28 23:18 GMT+02:00 Rui Barradas
> <ruipbarradas using sapo.pt<mailto:ruipbarradas using sapo.pt>>:
>
> I forgot to explain why my suggestion.
>
> The logical condition returns FALSE/TRUE that in R are coded as 0/1.
> So all you have to do is coerce to integer.
>
> This works because the ifelse will return a 1 or a 0 depending on
> the condition. Meaning exactly the same values. And is more
> efficient since ifelse creates both vectors, the true part and the
> false part, and then indexes those vectors in order to return the
> appropriate values. This is the double of the trouble and a great
> deal of memory used.
>
> Rui Barradas
>
> On 4/28/2018 10:12 PM, Rui Barradas wrote:
>
> Hello,
>
> instead of ifelse, the following is exactly the same and much
> more efficient.
>
> d0[[nm]] <- as.integer(regexpr(d1[i,1], d0$X0) > 0)
>
>
> Hope this helps,
>
> Rui Barradas
>
> On 4/28/2018 8:45 PM, Luca Meyer wrote:
>
> Thanks Don,
>
> for (i in 1:10){
> nm <- paste0("V", i)
> d0[[nm]] <- ifelse( regexpr(d1[i,1], d0$X0) > 0, 1, 0)
> }
>
> is exaclty what I needed.
>
> Best regards,
>
> Luca
>
>
> 2018-04-25 23:03 GMT+02:00 MacQueen, Don
> <macqueen1 using llnl.gov<mailto:macqueen1 using llnl.gov>>:
>
> Your code doesn't make sense to me in a couple of ways.
>
> Inside the loop, the first line assigns a value to an
> object named "t".
> Then, the second line does the same thing, assigns a
> value to an object
> named "t".
>
> The value of the object named "t" after the second line
> will be the output
> of the ifelse() expression, whatever that is. This has
> the effect of making
> the first line irrelevant. Whatever value t has after
> the first line is
> replaced by whatever it gets from the second line.
>
> It looks like the first line inside the loop is
> constructing the name of a
> data frame column, and storing that name as a character
> string. However,
> the second line doesn't use that name at all. If your
> goal is to update the
> contents of a column, you need to assign something to
> that column in the
> next line. Instead you assign it to the object named "t".
>
> What you're looking for will be more along the lines of
> this:
>
> for (i in 1:10){
> nm <- paste0("V", i)
> d0[[nm]] <- ifelse( regexpr(d1[i,1], d0$X0) > 0,
> 1, 0)
> }
>
> This may not a complete solution, since I have no idea
> what the contents
> or structure of d1 are, or what the regexpr() is
> expected to return.
>
> And notice the use of double brackets, [[ and ]]. This
> is one way to
> reference a column of a data frame when you have the
> column's name stored
> in a variable. Another way is d0[ , nm]
>
>
> A couple of additional comments:
>
> "t" is a poor choice of object name, because it is
> one of R's built-in
> functions (immediately after starting a fresh session of
> R, with nothing
> left over from any previous session, type help("r") and
> see what you get).
>
> ifelse() is intended for use on vectors, not scalars,
> and it looks like
> maybe you're using it on a scalar (can't be sure about
> this, though)
>
> For example, ifelse() is designed for this kind of usage:
>
> ifelse( c(TRUE, FALSE, TRUE) , 1:3, 11:13)
>
> [1] 1 12 3
>
> Although it works ok for these
>
> ifelse(TRUE, 3, 4)
>
> [1] 3
>
> ifelse(FALSE, 3, 4)
>
> [1] 4
> They are not really what it is intended for.
>
> --
> Don MacQueen
> Lawrence Livermore National Laboratory
> 7000 East Ave., L-627
> Livermore, CA 94550
> 925-423-1062
> Lab cell 925-724-7509
>
>
> On 4/24/18, 12:30 AM, "R-help on behalf of Luca Meyer" <
> r-help-bounces using r-project.org<mailto:r-help-bounces using r-project.org>on
> behalf of
> lucam1968 using gmail.com<mailto:lucam1968 using gmail.com>> wrote:
>
> Hi,
>
> I am trying to debug the following code:
>
> for (i in 1:10){
> t <- paste("d0$V",i,sep="")
> t <- ifelse(regexpr(d1[i,1],d0$X0)>0,1,0)
> }
>
> and I would like to see what code is actually
> processing R, how can I
> do
> that?
>
> More to the point, I am trying to update my
> variables d0$V1 to d0$V10
> according to the presence or absence of some text
> (contained in the
> file
> d1) within the d0$X0 variable.
>
> The code seem to run ok, if I add print(table(t))
> within the loop I
> can see
> that the ifelse procedure is working and to some
> cases within the
> d0$V1 to
> d0$V10 variable range a 1 is assigned. But when
> checking my d0$V1 to
> d0$V10
> after the for loop they are all still equal to zero...
>
> Thanks,
>
> Luca
>
> [[alternative HTML version deleted]]
>
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> [[alternative HTML version deleted]]
>
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> code.
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>
> ______________________________________________
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